CS 188: Artificial Intelligence Constraint Satisfaction Problems II Instructors: Dan Klein and Pieter Abbeel University of California, Berkeley [These slides were created by Dan Klein and Pieter Abbeel for CS188 Intro to AI at UC Berkeley. All CS188 materials are available at http://ai.berkeley.edu.]
Today Efficient Solution of CSPs Local Search
Reminder: CSPs CSPs: Variables Domains Constraints Implicit (provide code to compute) Explicit (provide a list of the legal tuples) Unary / Binary / N-ary Goals: Here: find any solution Also: find all, find best, etc.
Backtracking Search
Improving Backtracking General-purpose ideas give huge gains in speed … but it’s all still NP-hard Filtering: Can we detect inevitable failure early? Ordering: Which variable should be assigned next? (MRV) In what order should its values be tried? (LCV) Structure: Can we exploit the problem structure?
Arc Consistency and Beyond
Arc Consistency of an Entire CSP A simple form of propagation makes sure all arcs are simultaneously consistent: NT Q WA SA NSW V Arc consistency detects failure earlier than forward checking Remember: Delete Important: If X loses a value, neighbors of X need to be rechecked! from the tail! Must rerun after each assignment!
Limitations of Arc Consistency After enforcing arc consistency: Can have one solution left Can have multiple solutions left Can have no solutions left (and not know it) Arc consistency still runs What went wrong here? inside a backtracking search!
K-Consistency
K-Consistency Increasing degrees of consistency 1-Consistency (Node Consistency): Each single node’s domain has a value which meets that node’s unary constraints 2-Consistency (Arc Consistency): For each pair of nodes, any consistent assignment to one can be extended to the other K-Consistency: For each k nodes, any consistent assignment to k-1 can be extended to the k th node. Higher k more expensive to compute (You need to know the k=2 case: arc consistency)
Strong K-Consistency Strong k-consistency: also k-1, k-2, … 1 consistent Claim: strong n-consistency means we can solve without backtracking! Why? Choose any assignment to any variable Choose a new variable By 2-consistency, there is a choice consistent with the first Choose a new variable By 3-consistency, there is a choice consistent with the first 2 … Lots of middle ground between arc consistency and n-consistency! (e.g. k=3, called path consistency)
Structure
Problem Structure Extreme case: independent subproblems Example: Tasmania and mainland do not interact Independent subproblems are identifiable as connected components of constraint graph Suppose a graph of n variables can be broken into subproblems of only c variables: Worst-case solution cost is O((n/c)(d c )), linear in n E.g., n = 80, d = 2, c =20 2 80 = 4 billion years at 10 million nodes/sec (4)(2 20 ) = 0.4 seconds at 10 million nodes/sec
Tree-Structured CSPs Theorem: if the constraint graph has no loops, the CSP can be solved in O(n d 2 ) time Compare to general CSPs, where worst-case time is O(d n ) This property also applies to probabilistic reasoning (later): an example of the relation between syntactic restrictions and the complexity of reasoning
Tree-Structured CSPs Algorithm for tree-structured CSPs: Order: Choose a root variable, order variables so that parents precede children Remove backward: For i = n : 2, apply RemoveInconsistent(Parent(X i ),X i ) Assign forward: For i = 1 : n, assign X i consistently with Parent(X i ) Runtime: O(n d 2 ) (why?)
Tree-Structured CSPs Claim 1: After backward pass, all root-to-leaf arcs are consistent Proof: Each X → Y was made consistent at one point and Y’s domain could not have been reduced thereafter (because Y’s children were processed before Y) Claim 2: If root-to-leaf arcs are consistent, forward assignment will not backtrack Proof: Induction on position Why doesn’t this algorithm work with cycles in the constraint graph? Note: we’ll see this basic idea again with Bayes’ nets
Improving Structure
Nearly Tree-Structured CSPs Conditioning: instantiate a variable, prune its neighbors' domains Cutset conditioning: instantiate (in all ways) a set of variables such that the remaining constraint graph is a tree Cutset size c gives runtime O( (d c ) (n-c) d 2 ), very fast for small c
Cutset Conditioning Choose a cutset SA Instantiate the cutset (all possible ways) SA SA SA Compute residual CSP for each assignment Solve the residual CSPs (tree structured)
Cutset Quiz Find the smallest cutset for the graph below.
Tree Decomposition* Idea: create a tree-structured graph of mega-variables Each mega-variable encodes part of the original CSP Subproblems overlap to ensure consistent solutions M1 M2 M3 M4 ≠ ≠ ≠ ≠ Agree on shared vars Agree on shared vars Agree on shared vars NS NS WA NT NT Q Q V W W ≠ ≠ ≠ ≠ ≠ ≠ ≠ ≠ SA SA SA SA {(WA=r,SA=g,NT=b), {(NT=r,SA=g,Q=b), Agree: (M1,M2) ∈ (WA=b,SA=r,NT=g), (NT=b,SA=g,Q=r), {( (WA=g,SA=g,NT=g), (NT=g,SA=g,Q=g) ), …} …} …}
Iterative Improvement
Iterative Algorithms for CSPs Local search methods typically work with “complete” states, i.e., all variables assigned To apply to CSPs: Take an assignment with unsatisfied constraints Operators reassign variable values No fringe! Live on the edge. Algorithm: While not solved, Variable selection: randomly select any conflicted variable Value selection: min-conflicts heuristic: Choose a value that violates the fewest constraints I.e., hill climb with h(n) = total number of violated constraints
Example: 4-Queens States: 4 queens in 4 columns (4 4 = 256 states) Operators: move queen in column Goal test: no attacks Evaluation: c(n) = number of attacks [Demo: n-queens – iterative improvement (L5D1)] [Demo: coloring – iterative improvement]
Video of Demo Iterative Improvement – n Queens
Video of Demo Iterative Improvement – Coloring
Performance of Min-Conflicts Given random initial state, can solve n-queens in almost constant time for arbitrary n with high probability (e.g., n = 10,000,000)! The same appears to be true for any randomly-generated CSP except in a narrow range of the ratio
Summary: CSPs CSPs are a special kind of search problem: States are partial assignments Goal test defined by constraints Basic solution: backtracking search Speed-ups: Ordering Filtering Structure Iterative min-conflicts is often effective in practice
Local Search
Local Search Tree search keeps unexplored alternatives on the fringe (ensures completeness) Local search: improve a single option until you can’t make it better (no fringe!) New successor function: local changes Generally much faster and more memory efficient (but incomplete and suboptimal)
Hill Climbing Simple, general idea: Start wherever Repeat: move to the best neighboring state If no neighbors better than current, quit What’s bad about this approach? Complete? Optimal? What’s good about it?
Hill Climbing Diagram
Hill Climbing Quiz Starting from X, where do you end up ? Starting from Y, where do you end up ? Starting from Z, where do you end up ?
Simulated Annealing Idea: Escape local maxima by allowing downhill moves But make them rarer as time goes on 34
Simulated Annealing Theoretical guarantee: Stationary distribution: If T decreased slowly enough, will converge to optimal state! Is this an interesting guarantee? Sounds like magic, but reality is reality: The more downhill steps you need to escape a local optimum, the less likely you are to ever make them all in a row People think hard about ridge operators which let you jump around the space in better ways
Genetic Algorithms Genetic algorithms use a natural selection metaphor Keep best N hypotheses at each step (selection) based on a fitness function Also have pairwise crossover operators, with optional mutation to give variety Possibly the most misunderstood, misapplied (and even maligned) technique around
Example: N-Queens Why does crossover make sense here? When wouldn’t it make sense? What would mutation be? What would a good fitness function be?
Next Time: Adversarial Search!
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